Practice · Opposite signs turn |b|=2|a| into b=-2a · Sensim Math

Variant 1 easy

For two points X and Y on a number line, let X♥Y denote the coordinate of the midpoint of segment XY. For the two points A and B below, find the coordinate of point C, the point given by (A♥B)♥A.

Conditions, for the points A( $a$ ) and B( $b$ ):

(a) $2|a|=|b|$
(b) $a\times b<0$ and $a+b=-4$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Two points A( $a$ ) and B( $b$ ) satisfy $2|a| = |b|$ , $ab < 0$ and $a + b = -4$ . Find the midpoint of A and B, then the midpoint of that point and A.

Givens

$2|a| = |b|$
$a \times b < 0$
$a + b = -4$
X♥Y is the coordinate of the midpoint of segment XY.

Unknowns

The coordinate of C = (A♥B)♥A.

Constraints

Since $ab < 0$ , neither $a$ nor $b$ is $0$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The absolute-value condition alone allows $b = 2a$ or $b = -2a$ . The sign condition picks one, which turns the problem into a one-variable equation. After finding A and B, apply the midpoint rule twice on the number line.

3 · Execute4 carry out the plan

1Decide the signs

#13 Convert to Algebra 7.NS.A.2

A negative product means

a

and

b

have opposite signs.

|b| = 2|a|

says

b

is twice as far from

0

as

a

, and on the opposite side, so

b = -2a

.

a \times b < 0,\ |b| = 2|a| \Rightarrow b = -2a

b = 2a

would give

a

and

b

the same sign, which is ruled out.

2Use the sum

#13 Convert to Algebra 7.NS.A.1

Substitute

b = -2a

into

a + b = -4

.

a + (-2a) = -a = -4 \Rightarrow a = 4,\ b = -8

a = 4

and

b = -8

; check:

2 \times 4 = 8

,

4 \times (-8) < 0

,

4 + (-8) = -4

.

3First midpoint

#1 Draw a Diagram 7.NS.A.1

A♥B is the average of

4

and

-8

.

\frac{4 + (-8)}{2} = -2

The midpoint of A and B is at

-2

.

4Second midpoint

#7 Identify Subproblems 7.NS.A.1

(A♥B)♥A is the midpoint of the point

-2

and A(

4

).

\frac{-2 + 4}{2} = 1

Halfway between

-2

and

4

is

1

.

Answer:

1

4 · Reviewdoes it hold up?

On the number line, A is at $4$ and B at $-8$ , $12$ apart, so their midpoint is $6$ to the left of A, at $-2$ . The midpoint of $4$ and $-2$ is $3$ to the left of A, at $1$ . C lies a quarter of the way from A to B, as it should.

Another way: (A♥B)♥A $= \frac{\frac{a+b}{2} + a}{2} = \frac{3a + b}{4}$ , so with $a = 4$ , $b = -8$ it is $\frac{12 - 8}{4} = 1$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading $ab < 0$ as opposite signs.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed numbers to solve for $a$ and to find midpoints.
6.NS.C.7 Order and absolute value of rational numbers — Interpreting absolute value as distance from $0$.

💡Takeaway. A sign condition can turn an absolute-value equation with two cases into a single equation.

Variant 2 easy

For two points X and Y on a number line, let X♥Y denote the coordinate of the midpoint of segment XY. For the two points A and B below, find the coordinate of point C, the point given by (A♥B)♥A.

Conditions, for the points A( $a$ ) and B( $b$ ):

(a) $2|a|=|b|$
(b) $a\times b<0$ and $a+b=4$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Two points A( $a$ ) and B( $b$ ) satisfy $2|a| = |b|$ , $ab < 0$ and $a + b = 4$ . Find the midpoint of A and B, then the midpoint of that point and A.

Givens

$2|a| = |b|$
$a \times b < 0$
$a + b = 4$
X♥Y is the coordinate of the midpoint of segment XY.

Unknowns

The coordinate of C = (A♥B)♥A.

Constraints

Since $ab < 0$ , neither $a$ nor $b$ is $0$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The absolute-value condition alone allows $b = 2a$ or $b = -2a$ . The sign condition picks one, which turns the problem into a one-variable equation. After finding A and B, apply the midpoint rule twice on the number line.

3 · Execute4 carry out the plan

1Decide the signs

#13 Convert to Algebra 7.NS.A.2

A negative product means

a

and

b

have opposite signs.

|b| = 2|a|

says

b

is twice as far from

0

as

a

, and on the opposite side, so

b = -2a

.

a \times b < 0,\ |b| = 2|a| \Rightarrow b = -2a

b = 2a

would give

a

and

b

the same sign, which is ruled out.

2Use the sum

#13 Convert to Algebra 7.NS.A.1

Substitute

b = -2a

into

a + b = 4

.

a + (-2a) = -a = 4 \Rightarrow a = -4,\ b = 8

a = -4

and

b = 8

; check:

2 \times 4 = 8

,

(-4) \times 8 < 0

,

-4 + 8 = 4

.

3First midpoint

#1 Draw a Diagram 7.NS.A.1

A♥B is the average of

-4

and

8

.

\frac{(-4) + 8}{2} = 2

The midpoint of A and B is at

2

.

4Second midpoint

#7 Identify Subproblems 7.NS.A.1

(A♥B)♥A is the midpoint of the point

2

and A(

-4

).

\frac{2 + (-4)}{2} = -1

Halfway between

2

and

-4

is

-1

.

Answer:

-1

4 · Reviewdoes it hold up?

On the number line, A is at $-4$ and B at $8$ , $12$ apart, so their midpoint is $6$ to the right of A, at $2$ . The midpoint of $-4$ and $2$ is $3$ to the right of A, at $-1$ . C lies a quarter of the way from A to B, as it should.

Another way: (A♥B)♥A $= \frac{\frac{a+b}{2} + a}{2} = \frac{3a + b}{4}$ , so with $a = -4$ , $b = 8$ it is $\frac{-12 + 8}{4} = -1$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading $ab < 0$ as opposite signs.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed numbers to solve for $a$ and to find midpoints.
6.NS.C.7 Order and absolute value of rational numbers — Interpreting absolute value as distance from $0$.

💡Takeaway. A sign condition can turn an absolute-value equation with two cases into a single equation.

Variant 3 easy

For two points X and Y on a number line, let X♥Y denote the coordinate of the midpoint of segment XY. For the two points A and B below, find the coordinate of point C, the point given by (A♥B)♥A.

Conditions, for the points A( $a$ ) and B( $b$ ):

(a) $2|a|=|b|$
(b) $a\times b<0$ and $a+b=8$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Two points A( $a$ ) and B( $b$ ) satisfy $2|a| = |b|$ , $ab < 0$ and $a + b = 8$ . Find the midpoint of A and B, then the midpoint of that point and A.

Givens

$2|a| = |b|$
$a \times b < 0$
$a + b = 8$
X♥Y is the coordinate of the midpoint of segment XY.

Unknowns

The coordinate of C = (A♥B)♥A.

Constraints

Since $ab < 0$ , neither $a$ nor $b$ is $0$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The absolute-value condition alone allows $b = 2a$ or $b = -2a$ . The sign condition picks one, which turns the problem into a one-variable equation. After finding A and B, apply the midpoint rule twice on the number line.

3 · Execute4 carry out the plan

1Decide the signs

#13 Convert to Algebra 7.NS.A.2

A negative product means

a

and

b

have opposite signs.

|b| = 2|a|

says

b

is twice as far from

0

as

a

, and on the opposite side, so

b = -2a

.

a \times b < 0,\ |b| = 2|a| \Rightarrow b = -2a

b = 2a

would give

a

and

b

the same sign, which is ruled out.

2Use the sum

#13 Convert to Algebra 7.NS.A.1

Substitute

b = -2a

into

a + b = 8

.

a + (-2a) = -a = 8 \Rightarrow a = -8,\ b = 16

a = -8

and

b = 16

; check:

2 \times 8 = 16

,

(-8) \times 16 < 0

,

-8 + 16 = 8

.

3First midpoint

#1 Draw a Diagram 7.NS.A.1

A♥B is the average of

-8

and

16

.

\frac{(-8) + 16}{2} = 4

The midpoint of A and B is at

4

.

4Second midpoint

#7 Identify Subproblems 7.NS.A.1

(A♥B)♥A is the midpoint of the point

4

and A(

-8

).

\frac{4 + (-8)}{2} = -2

Halfway between

4

and

-8

is

-2

.

Answer:

-2

4 · Reviewdoes it hold up?

On the number line, A is at $-8$ and B at $16$ , $24$ apart, so their midpoint is $12$ to the right of A, at $4$ . The midpoint of $-8$ and $4$ is $6$ to the right of A, at $-2$ . C lies a quarter of the way from A to B, as it should.

Another way: (A♥B)♥A $= \frac{\frac{a+b}{2} + a}{2} = \frac{3a + b}{4}$ , so with $a = -8$ , $b = 16$ it is $\frac{-24 + 16}{4} = -2$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading $ab < 0$ as opposite signs.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed numbers to solve for $a$ and to find midpoints.
6.NS.C.7 Order and absolute value of rational numbers — Interpreting absolute value as distance from $0$.

💡Takeaway. A sign condition can turn an absolute-value equation with two cases into a single equation.

Variant 4 easy

For two points X and Y on a number line, let X♥Y denote the coordinate of the midpoint of segment XY. For the two points A and B below, find the coordinate of point C, the point given by (A♥B)♥A.

Conditions, for the points A( $a$ ) and B( $b$ ):

(a) $2|a|=|b|$
(b) $a\times b<0$ and $a+b=-8$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Two points A( $a$ ) and B( $b$ ) satisfy $2|a| = |b|$ , $ab < 0$ and $a + b = -8$ . Find the midpoint of A and B, then the midpoint of that point and A.

Givens

$2|a| = |b|$
$a \times b < 0$
$a + b = -8$
X♥Y is the coordinate of the midpoint of segment XY.

Unknowns

The coordinate of C = (A♥B)♥A.

Constraints

Since $ab < 0$ , neither $a$ nor $b$ is $0$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The absolute-value condition alone allows $b = 2a$ or $b = -2a$ . The sign condition picks one, which turns the problem into a one-variable equation. After finding A and B, apply the midpoint rule twice on the number line.

3 · Execute4 carry out the plan

1Decide the signs

#13 Convert to Algebra 7.NS.A.2

A negative product means

a

and

b

have opposite signs.

|b| = 2|a|

says

b

is twice as far from

0

as

a

, and on the opposite side, so

b = -2a

.

a \times b < 0,\ |b| = 2|a| \Rightarrow b = -2a

b = 2a

would give

a

and

b

the same sign, which is ruled out.

2Use the sum

#13 Convert to Algebra 7.NS.A.1

Substitute

b = -2a

into

a + b = -8

.

a + (-2a) = -a = -8 \Rightarrow a = 8,\ b = -16

a = 8

and

b = -16

; check:

2 \times 8 = 16

,

8 \times (-16) < 0

,

8 + (-16) = -8

.

3First midpoint

#1 Draw a Diagram 7.NS.A.1

A♥B is the average of

8

and

-16

.

\frac{8 + (-16)}{2} = -4

The midpoint of A and B is at

-4

.

4Second midpoint

#7 Identify Subproblems 7.NS.A.1

(A♥B)♥A is the midpoint of the point

-4

and A(

8

).

\frac{-4 + 8}{2} = 2

Halfway between

-4

and

8

is

2

.

Answer:

2

4 · Reviewdoes it hold up?

On the number line, A is at $8$ and B at $-16$ , $24$ apart, so their midpoint is $12$ to the left of A, at $-4$ . The midpoint of $8$ and $-4$ is $6$ to the left of A, at $2$ . C lies a quarter of the way from A to B, as it should.

Another way: (A♥B)♥A $= \frac{\frac{a+b}{2} + a}{2} = \frac{3a + b}{4}$ , so with $a = 8$ , $b = -16$ it is $\frac{24 - 16}{4} = 2$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading $ab < 0$ as opposite signs.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed numbers to solve for $a$ and to find midpoints.
6.NS.C.7 Order and absolute value of rational numbers — Interpreting absolute value as distance from $0$.

💡Takeaway. A sign condition can turn an absolute-value equation with two cases into a single equation.

Variant 5 medium

For two points X and Y on a number line, let X♥Y denote the coordinate of the midpoint of segment XY. For the two points A and B below, find the coordinate of point C, the point given by (A♥B)♥A.

Conditions, for the points A( $a$ ) and B( $b$ ):

(a) $5|a|=|b|$
(b) $a\times b<0$ and $a+b=-8$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Two points A( $a$ ) and B( $b$ ) satisfy $5|a| = |b|$ , $ab < 0$ and $a + b = -8$ . Find the midpoint of A and B, then the midpoint of that point and A.

Givens

$5|a| = |b|$
$a \times b < 0$
$a + b = -8$
X♥Y is the coordinate of the midpoint of segment XY.

Unknowns

The coordinate of C = (A♥B)♥A.

Constraints

Since $ab < 0$ , neither $a$ nor $b$ is $0$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The absolute-value condition alone allows $b = 5a$ or $b = -5a$ . The sign condition picks one, which turns the problem into a one-variable equation. After finding A and B, apply the midpoint rule twice on the number line.

3 · Execute4 carry out the plan

1Decide the signs

#13 Convert to Algebra 7.NS.A.2

A negative product means

a

and

b

have opposite signs.

|b| = 5|a|

says

b

is five times as far from

0

as

a

, and on the opposite side, so

b = -5a

.

a \times b < 0,\ |b| = 5|a| \Rightarrow b = -5a

b = 5a

would give

a

and

b

the same sign, which is ruled out.

2Use the sum

#13 Convert to Algebra 7.NS.A.1

Substitute

b = -5a

into

a + b = -8

.

a + (-5a) = -4a = -8 \Rightarrow a = 2,\ b = -10

a = 2

and

b = -10

; check:

5 \times 2 = 10

,

2 \times (-10) < 0

,

2 + (-10) = -8

.

3First midpoint

#1 Draw a Diagram 7.NS.A.1

A♥B is the average of

2

and

-10

.

\frac{2 + (-10)}{2} = -4

The midpoint of A and B is at

-4

.

4Second midpoint

#7 Identify Subproblems 7.NS.A.1

(A♥B)♥A is the midpoint of the point

-4

and A(

2

).

\frac{-4 + 2}{2} = -1

Halfway between

-4

and

2

is

-1

.

Answer:

-1

4 · Reviewdoes it hold up?

On the number line, A is at $2$ and B at $-10$ , $12$ apart, so their midpoint is $6$ to the left of A, at $-4$ . The midpoint of $2$ and $-4$ is $3$ to the left of A, at $-1$ . C lies a quarter of the way from A to B, as it should.

Another way: (A♥B)♥A $= \frac{\frac{a+b}{2} + a}{2} = \frac{3a + b}{4}$ , so with $a = 2$ , $b = -10$ it is $\frac{6 - 10}{4} = -1$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading $ab < 0$ as opposite signs.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed numbers to solve for $a$ and to find midpoints.
6.NS.C.7 Order and absolute value of rational numbers — Interpreting absolute value as distance from $0$.

💡Takeaway. A sign condition can turn an absolute-value equation with two cases into a single equation.

Variant 6 medium

For two points X and Y on a number line, let X♥Y denote the coordinate of the midpoint of segment XY. For the two points A and B below, find the coordinate of point C, the point given by (A♥B)♥A.

Conditions, for the points A( $a$ ) and B( $b$ ):

(a) $5|a|=|b|$
(b) $a\times b<0$ and $a+b=8$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Two points A( $a$ ) and B( $b$ ) satisfy $5|a| = |b|$ , $ab < 0$ and $a + b = 8$ . Find the midpoint of A and B, then the midpoint of that point and A.

Givens

$5|a| = |b|$
$a \times b < 0$
$a + b = 8$
X♥Y is the coordinate of the midpoint of segment XY.

Unknowns

The coordinate of C = (A♥B)♥A.

Constraints

Since $ab < 0$ , neither $a$ nor $b$ is $0$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The absolute-value condition alone allows $b = 5a$ or $b = -5a$ . The sign condition picks one, which turns the problem into a one-variable equation. After finding A and B, apply the midpoint rule twice on the number line.

3 · Execute4 carry out the plan

1Decide the signs

#13 Convert to Algebra 7.NS.A.2

A negative product means

a

and

b

have opposite signs.

|b| = 5|a|

says

b

is five times as far from

0

as

a

, and on the opposite side, so

b = -5a

.

a \times b < 0,\ |b| = 5|a| \Rightarrow b = -5a

b = 5a

would give

a

and

b

the same sign, which is ruled out.

2Use the sum

#13 Convert to Algebra 7.NS.A.1

Substitute

b = -5a

into

a + b = 8

.

a + (-5a) = -4a = 8 \Rightarrow a = -2,\ b = 10

a = -2

and

b = 10

; check:

5 \times 2 = 10

,

(-2) \times 10 < 0

,

-2 + 10 = 8

.

3First midpoint

#1 Draw a Diagram 7.NS.A.1

A♥B is the average of

-2

and

10

.

\frac{(-2) + 10}{2} = 4

The midpoint of A and B is at

4

.

4Second midpoint

#7 Identify Subproblems 7.NS.A.1

(A♥B)♥A is the midpoint of the point

4

and A(

-2

).

\frac{4 + (-2)}{2} = 1

Halfway between

4

and

-2

is

1

.

Answer:

1

4 · Reviewdoes it hold up?

On the number line, A is at $-2$ and B at $10$ , $12$ apart, so their midpoint is $6$ to the right of A, at $4$ . The midpoint of $-2$ and $4$ is $3$ to the right of A, at $1$ . C lies a quarter of the way from A to B, as it should.

Another way: (A♥B)♥A $= \frac{\frac{a+b}{2} + a}{2} = \frac{3a + b}{4}$ , so with $a = -2$ , $b = 10$ it is $\frac{-6 + 10}{4} = 1$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading $ab < 0$ as opposite signs.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed numbers to solve for $a$ and to find midpoints.
6.NS.C.7 Order and absolute value of rational numbers — Interpreting absolute value as distance from $0$.

💡Takeaway. A sign condition can turn an absolute-value equation with two cases into a single equation.

Variant 7 hard

For two points X and Y on a number line, let X♥Y denote the coordinate of the midpoint of segment XY. For the two points A and B below, find the coordinate of point C, the point given by (A♥B)♥A.

Conditions, for the points A( $a$ ) and B( $b$ ):

(a) $4|a|=|b|$
(b) $a\times b<0$ and $a+b=-12$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Two points A( $a$ ) and B( $b$ ) satisfy $4|a| = |b|$ , $ab < 0$ and $a + b = -12$ . Find the midpoint of A and B, then the midpoint of that point and A.

Givens

$4|a| = |b|$
$a \times b < 0$
$a + b = -12$
X♥Y is the coordinate of the midpoint of segment XY.

Unknowns

The coordinate of C = (A♥B)♥A.

Constraints

Since $ab < 0$ , neither $a$ nor $b$ is $0$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The absolute-value condition alone allows $b = 4a$ or $b = -4a$ . The sign condition picks one, which turns the problem into a one-variable equation. After finding A and B, apply the midpoint rule twice on the number line.

3 · Execute4 carry out the plan

1Decide the signs

#13 Convert to Algebra 7.NS.A.2

A negative product means

a

and

b

have opposite signs.

|b| = 4|a|

says

b

is four times as far from

0

as

a

, and on the opposite side, so

b = -4a

.

a \times b < 0,\ |b| = 4|a| \Rightarrow b = -4a

b = 4a

would give

a

and

b

the same sign, which is ruled out.

2Use the sum

#13 Convert to Algebra 7.NS.A.1

Substitute

b = -4a

into

a + b = -12

.

a + (-4a) = -3a = -12 \Rightarrow a = 4,\ b = -16

a = 4

and

b = -16

; check:

4 \times 4 = 16

,

4 \times (-16) < 0

,

4 + (-16) = -12

.

3First midpoint

#1 Draw a Diagram 7.NS.A.1

A♥B is the average of

4

and

-16

.

\frac{4 + (-16)}{2} = -6

The midpoint of A and B is at

-6

.

4Second midpoint

#7 Identify Subproblems 7.NS.A.1

(A♥B)♥A is the midpoint of the point

-6

and A(

4

).

\frac{-6 + 4}{2} = -1

Halfway between

-6

and

4

is

-1

.

Answer:

-1

4 · Reviewdoes it hold up?

On the number line, A is at $4$ and B at $-16$ , $20$ apart, so their midpoint is $10$ to the left of A, at $-6$ . The midpoint of $4$ and $-6$ is $5$ to the left of A, at $-1$ . C lies a quarter of the way from A to B, as it should.

Another way: (A♥B)♥A $= \frac{\frac{a+b}{2} + a}{2} = \frac{3a + b}{4}$ , so with $a = 4$ , $b = -16$ it is $\frac{12 - 16}{4} = -1$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading $ab < 0$ as opposite signs.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed numbers to solve for $a$ and to find midpoints.
6.NS.C.7 Order and absolute value of rational numbers — Interpreting absolute value as distance from $0$.

💡Takeaway. A sign condition can turn an absolute-value equation with two cases into a single equation.

Variant 8 medium

For two points X and Y on a number line, let X♥Y denote the coordinate of the midpoint of segment XY. For the two points A and B below, find the coordinate of point C, the point given by (A♥B)♥A.

Conditions, for the points A( $a$ ) and B( $b$ ):

(a) $2|a|=|b|$
(b) $a\times b<0$ and $a+b=12$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Two points A( $a$ ) and B( $b$ ) satisfy $2|a| = |b|$ , $ab < 0$ and $a + b = 12$ . Find the midpoint of A and B, then the midpoint of that point and A.

Givens

$2|a| = |b|$
$a \times b < 0$
$a + b = 12$
X♥Y is the coordinate of the midpoint of segment XY.

Unknowns

The coordinate of C = (A♥B)♥A.

Constraints

Since $ab < 0$ , neither $a$ nor $b$ is $0$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The absolute-value condition alone allows $b = 2a$ or $b = -2a$ . The sign condition picks one, which turns the problem into a one-variable equation. After finding A and B, apply the midpoint rule twice on the number line.

3 · Execute4 carry out the plan

1Decide the signs

#13 Convert to Algebra 7.NS.A.2

A negative product means

a

and

b

have opposite signs.

|b| = 2|a|

says

b

is twice as far from

0

as

a

, and on the opposite side, so

b = -2a

.

a \times b < 0,\ |b| = 2|a| \Rightarrow b = -2a

b = 2a

would give

a

and

b

the same sign, which is ruled out.

2Use the sum

#13 Convert to Algebra 7.NS.A.1

Substitute

b = -2a

into

a + b = 12

.

a + (-2a) = -a = 12 \Rightarrow a = -12,\ b = 24

a = -12

and

b = 24

; check:

2 \times 12 = 24

,

(-12) \times 24 < 0

,

-12 + 24 = 12

.

3First midpoint

#1 Draw a Diagram 7.NS.A.1

A♥B is the average of

-12

and

24

.

\frac{(-12) + 24}{2} = 6

The midpoint of A and B is at

6

.

4Second midpoint

#7 Identify Subproblems 7.NS.A.1

(A♥B)♥A is the midpoint of the point

6

and A(

-12

).

\frac{6 + (-12)}{2} = -3

Halfway between

6

and

-12

is

-3

.

Answer:

-3

4 · Reviewdoes it hold up?

On the number line, A is at $-12$ and B at $24$ , $36$ apart, so their midpoint is $18$ to the right of A, at $6$ . The midpoint of $-12$ and $6$ is $9$ to the right of A, at $-3$ . C lies a quarter of the way from A to B, as it should.

Another way: (A♥B)♥A $= \frac{\frac{a+b}{2} + a}{2} = \frac{3a + b}{4}$ , so with $a = -12$ , $b = 24$ it is $\frac{-36 + 24}{4} = -3$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading $ab < 0$ as opposite signs.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed numbers to solve for $a$ and to find midpoints.
6.NS.C.7 Order and absolute value of rational numbers — Interpreting absolute value as distance from $0$.

💡Takeaway. A sign condition can turn an absolute-value equation with two cases into a single equation.

Variant 9 medium

For two points X and Y on a number line, let X♥Y denote the coordinate of the midpoint of segment XY. For the two points A and B below, find the coordinate of point C, the point given by (A♥B)♥A.

Conditions, for the points A( $a$ ) and B( $b$ ):

(a) $4|a|=|b|$
(b) $a\times b<0$ and $a+b=12$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Two points A( $a$ ) and B( $b$ ) satisfy $4|a| = |b|$ , $ab < 0$ and $a + b = 12$ . Find the midpoint of A and B, then the midpoint of that point and A.

Givens

$4|a| = |b|$
$a \times b < 0$
$a + b = 12$
X♥Y is the coordinate of the midpoint of segment XY.

Unknowns

The coordinate of C = (A♥B)♥A.

Constraints

Since $ab < 0$ , neither $a$ nor $b$ is $0$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The absolute-value condition alone allows $b = 4a$ or $b = -4a$ . The sign condition picks one, which turns the problem into a one-variable equation. After finding A and B, apply the midpoint rule twice on the number line.

3 · Execute4 carry out the plan

1Decide the signs

#13 Convert to Algebra 7.NS.A.2

A negative product means

a

and

b

have opposite signs.

|b| = 4|a|

says

b

is four times as far from

0

as

a

, and on the opposite side, so

b = -4a

.

a \times b < 0,\ |b| = 4|a| \Rightarrow b = -4a

b = 4a

would give

a

and

b

the same sign, which is ruled out.

2Use the sum

#13 Convert to Algebra 7.NS.A.1

Substitute

b = -4a

into

a + b = 12

.

a + (-4a) = -3a = 12 \Rightarrow a = -4,\ b = 16

a = -4

and

b = 16

; check:

4 \times 4 = 16

,

(-4) \times 16 < 0

,

-4 + 16 = 12

.

3First midpoint

#1 Draw a Diagram 7.NS.A.1

A♥B is the average of

-4

and

16

.

\frac{(-4) + 16}{2} = 6

The midpoint of A and B is at

6

.

4Second midpoint

#7 Identify Subproblems 7.NS.A.1

(A♥B)♥A is the midpoint of the point

6

and A(

-4

).

\frac{6 + (-4)}{2} = 1

Halfway between

6

and

-4

is

1

.

Answer:

1

4 · Reviewdoes it hold up?

On the number line, A is at $-4$ and B at $16$ , $20$ apart, so their midpoint is $10$ to the right of A, at $6$ . The midpoint of $-4$ and $6$ is $5$ to the right of A, at $1$ . C lies a quarter of the way from A to B, as it should.

Another way: (A♥B)♥A $= \frac{\frac{a+b}{2} + a}{2} = \frac{3a + b}{4}$ , so with $a = -4$ , $b = 16$ it is $\frac{-12 + 16}{4} = 1$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading $ab < 0$ as opposite signs.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed numbers to solve for $a$ and to find midpoints.
6.NS.C.7 Order and absolute value of rational numbers — Interpreting absolute value as distance from $0$.

💡Takeaway. A sign condition can turn an absolute-value equation with two cases into a single equation.

Variant 10 hard

For two points X and Y on a number line, let X♥Y denote the coordinate of the midpoint of segment XY. For the two points A and B below, find the coordinate of point C, the point given by (A♥B)♥A.

Conditions, for the points A( $a$ ) and B( $b$ ):

(a) $2|a|=|b|$
(b) $a\times b<0$ and $a+b=-16$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Two points A( $a$ ) and B( $b$ ) satisfy $2|a| = |b|$ , $ab < 0$ and $a + b = -16$ . Find the midpoint of A and B, then the midpoint of that point and A.

Givens

$2|a| = |b|$
$a \times b < 0$
$a + b = -16$
X♥Y is the coordinate of the midpoint of segment XY.

Unknowns

The coordinate of C = (A♥B)♥A.

Constraints

Since $ab < 0$ , neither $a$ nor $b$ is $0$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The absolute-value condition alone allows $b = 2a$ or $b = -2a$ . The sign condition picks one, which turns the problem into a one-variable equation. After finding A and B, apply the midpoint rule twice on the number line.

3 · Execute4 carry out the plan

1Decide the signs

#13 Convert to Algebra 7.NS.A.2

A negative product means

a

and

b

have opposite signs.

|b| = 2|a|

says

b

is twice as far from

0

as

a

, and on the opposite side, so

b = -2a

.

a \times b < 0,\ |b| = 2|a| \Rightarrow b = -2a

b = 2a

would give

a

and

b

the same sign, which is ruled out.

2Use the sum

#13 Convert to Algebra 7.NS.A.1

Substitute

b = -2a

into

a + b = -16

.

a + (-2a) = -a = -16 \Rightarrow a = 16,\ b = -32

a = 16

and

b = -32

; check:

2 \times 16 = 32

,

16 \times (-32) < 0

,

16 + (-32) = -16

.

3First midpoint

#1 Draw a Diagram 7.NS.A.1

A♥B is the average of

16

and

-32

.

\frac{16 + (-32)}{2} = -8

The midpoint of A and B is at

-8

.

4Second midpoint

#7 Identify Subproblems 7.NS.A.1

(A♥B)♥A is the midpoint of the point

-8

and A(

16

).

\frac{-8 + 16}{2} = 4

Halfway between

-8

and

16

is

4

.

Answer:

4

4 · Reviewdoes it hold up?

On the number line, A is at $16$ and B at $-32$ , $48$ apart, so their midpoint is $24$ to the left of A, at $-8$ . The midpoint of $16$ and $-8$ is $12$ to the left of A, at $4$ . C lies a quarter of the way from A to B, as it should.

Another way: (A♥B)♥A $= \frac{\frac{a+b}{2} + a}{2} = \frac{3a + b}{4}$ , so with $a = 16$ , $b = -32$ it is $\frac{48 - 32}{4} = 4$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading $ab < 0$ as opposite signs.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed numbers to solve for $a$ and to find midpoints.
6.NS.C.7 Order and absolute value of rational numbers — Interpreting absolute value as distance from $0$.

💡Takeaway. A sign condition can turn an absolute-value equation with two cases into a single equation.

Variant 11 hard

For two points X and Y on a number line, let X♥Y denote the coordinate of the midpoint of segment XY. For the two points A and B below, find the coordinate of point C, the point given by (A♥B)♥A.

Conditions, for the points A( $a$ ) and B( $b$ ):

(a) $5|a|=|b|$
(b) $a\times b<0$ and $a+b=16$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Two points A( $a$ ) and B( $b$ ) satisfy $5|a| = |b|$ , $ab < 0$ and $a + b = 16$ . Find the midpoint of A and B, then the midpoint of that point and A.

Givens

$5|a| = |b|$
$a \times b < 0$
$a + b = 16$
X♥Y is the coordinate of the midpoint of segment XY.

Unknowns

The coordinate of C = (A♥B)♥A.

Constraints

Since $ab < 0$ , neither $a$ nor $b$ is $0$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The absolute-value condition alone allows $b = 5a$ or $b = -5a$ . The sign condition picks one, which turns the problem into a one-variable equation. After finding A and B, apply the midpoint rule twice on the number line.

3 · Execute4 carry out the plan

1Decide the signs

#13 Convert to Algebra 7.NS.A.2

A negative product means

a

and

b

have opposite signs.

|b| = 5|a|

says

b

is five times as far from

0

as

a

, and on the opposite side, so

b = -5a

.

a \times b < 0,\ |b| = 5|a| \Rightarrow b = -5a

b = 5a

would give

a

and

b

the same sign, which is ruled out.

2Use the sum

#13 Convert to Algebra 7.NS.A.1

Substitute

b = -5a

into

a + b = 16

.

a + (-5a) = -4a = 16 \Rightarrow a = -4,\ b = 20

a = -4

and

b = 20

; check:

5 \times 4 = 20

,

(-4) \times 20 < 0

,

-4 + 20 = 16

.

3First midpoint

#1 Draw a Diagram 7.NS.A.1

A♥B is the average of

-4

and

20

.

\frac{(-4) + 20}{2} = 8

The midpoint of A and B is at

8

.

4Second midpoint

#7 Identify Subproblems 7.NS.A.1

(A♥B)♥A is the midpoint of the point

8

and A(

-4

).

\frac{8 + (-4)}{2} = 2

Halfway between

8

and

-4

is

2

.

Answer:

2

4 · Reviewdoes it hold up?

On the number line, A is at $-4$ and B at $20$ , $24$ apart, so their midpoint is $12$ to the right of A, at $8$ . The midpoint of $-4$ and $8$ is $6$ to the right of A, at $2$ . C lies a quarter of the way from A to B, as it should.

Another way: (A♥B)♥A $= \frac{\frac{a+b}{2} + a}{2} = \frac{3a + b}{4}$ , so with $a = -4$ , $b = 20$ it is $\frac{-12 + 20}{4} = 2$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading $ab < 0$ as opposite signs.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed numbers to solve for $a$ and to find midpoints.
6.NS.C.7 Order and absolute value of rational numbers — Interpreting absolute value as distance from $0$.

💡Takeaway. A sign condition can turn an absolute-value equation with two cases into a single equation.

Variant 12 hard

For two points X and Y on a number line, let X♥Y denote the coordinate of the midpoint of segment XY. For the two points A and B below, find the coordinate of point C, the point given by (A♥B)♥A.

Conditions, for the points A( $a$ ) and B( $b$ ):

(a) $4|a|=|b|$
(b) $a\times b<0$ and $a+b=24$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Two points A( $a$ ) and B( $b$ ) satisfy $4|a| = |b|$ , $ab < 0$ and $a + b = 24$ . Find the midpoint of A and B, then the midpoint of that point and A.

Givens

$4|a| = |b|$
$a \times b < 0$
$a + b = 24$
X♥Y is the coordinate of the midpoint of segment XY.

Unknowns

The coordinate of C = (A♥B)♥A.

Constraints

Since $ab < 0$ , neither $a$ nor $b$ is $0$ .

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The absolute-value condition alone allows $b = 4a$ or $b = -4a$ . The sign condition picks one, which turns the problem into a one-variable equation. After finding A and B, apply the midpoint rule twice on the number line.

3 · Execute4 carry out the plan

1Decide the signs

#13 Convert to Algebra 7.NS.A.2

A negative product means

a

and

b

have opposite signs.

|b| = 4|a|

says

b

is four times as far from

0

as

a

, and on the opposite side, so

b = -4a

.

a \times b < 0,\ |b| = 4|a| \Rightarrow b = -4a

b = 4a

would give

a

and

b

the same sign, which is ruled out.

2Use the sum

#13 Convert to Algebra 7.NS.A.1

Substitute

b = -4a

into

a + b = 24

.

a + (-4a) = -3a = 24 \Rightarrow a = -8,\ b = 32

a = -8

and

b = 32

; check:

4 \times 8 = 32

,

(-8) \times 32 < 0

,

-8 + 32 = 24

.

3First midpoint

#1 Draw a Diagram 7.NS.A.1

A♥B is the average of

-8

and

32

.

\frac{(-8) + 32}{2} = 12

The midpoint of A and B is at

12

.

4Second midpoint

#7 Identify Subproblems 7.NS.A.1

(A♥B)♥A is the midpoint of the point

12

and A(

-8

).

\frac{12 + (-8)}{2} = 2

Halfway between

12

and

-8

is

2

.

Answer:

2

4 · Reviewdoes it hold up?

On the number line, A is at $-8$ and B at $32$ , $40$ apart, so their midpoint is $20$ to the right of A, at $12$ . The midpoint of $-8$ and $12$ is $10$ to the right of A, at $2$ . C lies a quarter of the way from A to B, as it should.

Another way: (A♥B)♥A $= \frac{\frac{a+b}{2} + a}{2} = \frac{3a + b}{4}$ , so with $a = -8$ , $b = 32$ it is $\frac{-24 + 32}{4} = 2$ .

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading $ab < 0$ as opposite signs.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Adding signed numbers to solve for $a$ and to find midpoints.
6.NS.C.7 Order and absolute value of rational numbers — Interpreting absolute value as distance from $0$.

💡Takeaway. A sign condition can turn an absolute-value equation with two cases into a single equation.

Opposite signs turn |b|=2|a| into b=-2a · 12 practice problems

Generated variants — 12

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Unknowns

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Unknowns

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Givens

Unknowns

Constraints

Givens

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Givens

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